A130249 Maximal index k of a Jacobsthal number such that A001045(k)<=n (the 'lower' Jacobsthal inverse).

0, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8

Offset

0,2

Comments

Inverse of the Jacobsthal sequence (A001045), nearly, since a(A001045(n))=n except for n=1 (see A130250 for another version). a(n)+1 is equal to the partial sum of the Jacobsthal indicator sequence (see A105348).

Links

G. C. Greubel, Table of n, a(n) for n = 0..10000

Formula

a(n) = floor(log_2(3n+1)).

a(n) = A130250(n+1) - 1 = A130253(n) - 1.

G.f.: 1/(1-x)*(Sum_{k>=1} x^A001045(k)).

Example

a(12)=5, since A001045(5)=11<=12, but A001045(6)=21>12.

Mathematica

Table[Floor[Log[2, 3*n + 1]], {n, 0, 50}] (* G. C. Greubel, Jan 08 2018 *)

Prog

(PARI) for(n=0, 30, print1(floor(log(3*n+1)/log(2)), ", ")) \\ G. C. Greubel, Jan 08 2018
(Magma) [Floor(Log(3*n+1)/Log(2)): n in [0..30]]; // G. C. Greubel, Jan 08 2018
(Python)
def A130249(n): return (3*n+1).bit_length()-1 # Chai Wah Wu, Jun 08 2022

Crossrefs

For partial sums see A130251.

Other related sequences A130250, A130253, A105348. A001045, A130233, A130241.

Sequence in context: A341166 A274102 A261100 * A286105 A061071 A122258

Adjacent sequences: A130246 A130247 A130248 * A130250 A130251 A130252

Keyword

nonn

Author

Hieronymus Fischer, May 20 2007

Status

approved